(A) The total number of moles of $K^+$ ions is calculated as follows:$n(K^+) = n(KCl) + 2 \times n(K_2SO_4)$$n(K^+) = (M_1 \times V_1) + 2 \times (M_2

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(A) The total number of moles of $K^+$ ions is calculated as follows:$n(K^+) = n(KCl) + 2 \times n(K_2SO_4)$$n(K^+) = (M_1 \times V_1) + 2 \times (M_2 \times V_2)$$n(K^+) = (2 \times 0.1) + 2 \times (3 \times 0.2) = 0.2 + 1.2 = 1.4 \ mol$The total volume of the mixture is $V_{total} = 100 \ mL + 200 \ mL = 300 \ mL = 0.3 \ L$The molarity of $K^+$ ions is $[K^+] = \frac{n(K^+)}{V_{total}} = \frac{1.4 \ mol}{0.3 \ L} = 4.667 \ M$

Class 12 Chemistry Solutions Method of expressing concentration of solution

Easy

When $100 \ mL$ of $2 \ M \ KCl$ and $200 \ mL$ of $3 \ M \ K_2SO_4$ solutions are mixed,what will be the molarity of $K^+$ ions in the resulting mixture (in $M$)?

A
$4.667$
B
$5$
C
$8$
D
$3.67$

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When $100 \ mL$ of $2 \ M \ KCl$ and $200 \ mL$ of $3 \ M \ K_2SO_4$ solutions are mixed,what will be the molarity of $K^+$ ions in the resulting mixture (in $M$)?

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Which of the following properties will change when the system containing solution $1$ becomes solution $2$?

$10 \ mol$ of solute $x + 10 \ L$ of water $1 \ L$ of solution $1 + 1 \ mol$ of solute $x + 1 \ L$ of water

(Solution $1$) (Solution $2$)